如何在 Laravel 中将 css 嵌入 HTML？
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How to embed css to HTML in laravel?
提问by cytsunny
First of all, to make things clear, I know I can include my css file by
首先，为了清楚起见，我知道我可以通过以下方式包含我的 css 文件
{{ HTML::style('css/myStyle.css') }}
Which the html to be generated will be:
要生成的 html 将是：
<link rel="stylesheet" type="text/css" href="http://myhost/public/css/myStyle.css">
The problem is that my view file is used for email and many email client don't load external css file. So I need to make it something like:
问题是我的视图文件用于电子邮件，许多电子邮件客户端不加载外部 css 文件。所以我需要使它像：
<style>
.content-of-my-css {
//content of my css
}
// more css
</style>
As the css settings will be used for multiple email template, I don't want to hard code it. In pure PHP, this can be done by (reference)
由于 css 设置将用于多个电子邮件模板，因此我不想对其进行硬编码。在纯 PHP 中，这可以通过（参考）完成
<style><?php include "path/to/my/css";?></style>
However, when I do that with laravel, it says:
但是，当我用 laravel 这样做时，它说：
ErrorException (E_ERROR) include(): http:// wrapper is disabled in the server configuration by allow_url_include=0
ErrorException (E_ERROR) include(): http://wrapper is disabled in the server configuration by allow_url_include=0
So, is there a way I can embed the css in laravel? (I am using laravel 4.2)
那么，有没有办法将 css 嵌入到 Laravel 中？（我使用的是laravel 4.2）
回答by GeorgeQ
It appears that you are using a URL instead of the local file path.
看来您使用的是 URL 而不是本地文件路径。
For example
例如
include('http://example.com/css/styles.css');
will produce your error but the following should not
会产生你的错误，但以下不应该
include(public_path().'css/styles.css');
回答by umefarooq
if you are using multiple template i will suggest you to create multiple style views which contain just css and you can include them with blade @include
如果您使用多个模板，我会建议您创建多个仅包含 css 的样式视图，您可以使用刀片 @include 包含它们
style1.blade.php
style2.blade.php
in styles just put css code
在样式中只需放置 css 代码
<style>
.content-of-my-css {
//content of my css
}
// more css
</style>
in master.blade.php layout
在 master.blade.php 布局中
<html>
<head>@include('style1')</head>
</html>
if it is dynamic you can pass style parameter
如果它是动态的，您可以传递样式参数
@include($style)
